Results 1 to 5 of 5

Thread: Arclength of linear curve

  1. #1
    Member
    Joined
    Oct 2009
    Posts
    196
    Thanks
    1

    Arclength of linear curve

    I'm to calculate the line segment y=mx+b between (x0, y0) and (x1, y1), using the formula for arclength when y is a function of x, that is:

    L = (integral evaluated at a and b) (root)(1 + (f'(x))^2

    Which in this case should give:

    L = (integral evaluated at x0 and x1) (root) (1+m^2), as m is the derivative of mx+b.

    This I get to be [(root)(1+m^2)*t]evaluated at x0 and x1, but the answer in my textbook is (1+m^2)|x0-x1|... any help would be appreciated! Thanks!
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    Joined
    Mar 2010
    From
    Florida
    Posts
    3,093
    Thanks
    10
    I am not sure about your books answer since m is just a constant.

    \int_{x_0}^{x_1}\sqrt{1+m^2}dx

    \sqrt{1+m^2} this is just some arbitrary number so move outside the integral.

    =\sqrt{1+m^2}*\int_{x_0}^{x_1}dx

    =\sqrt{1+m^2}*\big[x\big]_{x_0}^{x_1}

    =\sqrt{1+m^2}*(x_1-x_0)

    This what I obtained as a solution to the definite integral.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    MHF Contributor
    skeeter's Avatar
    Joined
    Jun 2008
    From
    North Texas
    Posts
    16,072
    Thanks
    3634
    Quote Originally Posted by gralla55 View Post
    I'm to calculate the line segment y=mx+b between (x0, y0) and (x1, y1), using the formula for arclength when y is a function of x, that is:

    L = (integral evaluated at a and b) (root)(1 + (f'(x))^2

    Which in this case should give:

    L = (integral evaluated at x0 and x1) (root) (1+m^2), as m is the derivative of mx+b.

    This I get to be [(root)(1+m^2)*t]evaluated at x0 and x1, but the answer in my textbook is (1+m^2)|x0-x1|... any help would be appreciated! Thanks!
    remember that you're setting up a general case arc length integral.

    what if x_1 < x_0 ?
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor
    Joined
    Mar 2010
    From
    Florida
    Posts
    3,093
    Thanks
    10
    Skeeter the book also got rid of the radical unless he forgot to type it.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Member
    Joined
    Oct 2009
    Posts
    196
    Thanks
    1
    Yes, the missing root sign is what confused me... but I guess the book could be wrong. Thanks anyway!
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Arclength?
    Posted in the Calculus Forum
    Replies: 4
    Last Post: Mar 22nd 2011, 03:04 PM
  2. Linear region of a sigmoidal curve
    Posted in the Algebra Forum
    Replies: 7
    Last Post: Nov 12th 2010, 08:08 AM
  3. help me with arclength
    Posted in the Calculus Forum
    Replies: 4
    Last Post: Oct 10th 2009, 03:43 PM
  4. help with arclength please
    Posted in the Calculus Forum
    Replies: 8
    Last Post: Sep 21st 2009, 03:41 PM
  5. [SOLVED] [SOLVED] Finding the arclength of a curve
    Posted in the Calculus Forum
    Replies: 2
    Last Post: Feb 25th 2008, 05:45 AM

Search Tags


/mathhelpforum @mathhelpforum